Download Math 396. Bijectivity vs. isomorphism 1. Motivation Let f : X → Y be a

Math 396. Bijectivity vs. isomorphism
1. Motivation
Let f : X → Y be a C p map between two C p -premanifolds with corners, with 1 ≤ p ≤ ∞.
Assuming f is bijective, we would like a criterion to tell us that f −1 is a C p map as well (so f is a
C p isomorphism and f (∂X) = ∂Y ). However, this can fail even if f −1 is continuous. For example,
the map f : R → R defined by f (t) = t3 is a C ∞ bijection, but its inverse t 7→ t1/3 is continuous
but is not differentiable at the origin. One necessary condition is that f should be an immersion.
In pathological situations, the “bijective immersion” condition is still not enough: if we let Y = R
with its usual topology and usual C ∞ -structure, and we let X = R with the discrete topology and
the associated C ∞ -structure, the identity map X → Y is a C ∞ bijective immersion but the inverse
is not even continuous. This latter example is pathological because R with the discrete topology
is not a second-countable space (as an uncountable discrete set cannot have a countable base of
opens, as all points are open and there are uncountably many of them). In a sense, the problem is
that the dimensions of the two spaces are not the same.
Let us now restrict ourselves to the second-countable case, as this prevents us from doing silly
things like making topologies discrete. But even with second-countability and Hausdorff conditions,
we still run into problems. For example, let Y = R (with its usual C ∞ -structure) and let X be
the disjoint union of R − {0} (with its usual C ∞ -structure) and the point {0} that we declare to
be open. There is an evident C ∞ bijection X → Y that is also an immersion, but once again the
inverse is not even continuous. Here the problem is not that the pointwise dimensions on X are all
too small, but that they “drop” somewhere. The same sort of example could be made much more
generally, such as taking Y to be open in Rn and X to be the disjoint union of a hyperplane slice
of Y (considered as an open set in an (n − 1)-dimensional vector space) and the open complement
of this hyperplane slice in Y .
Even when dimensions do not drop, there remains one further complication when our spaces have
corners. It is best illustrated with an example: f : [0, 1) → S 1 defined by t 7→ (cos(2πt), sin(2πt))
is a bijective immersion between connected 1-dimensional manifolds with corners, but the inverse
is not continuous. Here, the problem is that f (∂X) (a point) is not contained in ∂Y (the empty
set). Such a containment was a condition in the inverse function theorem on open sets in sectors,
so we should expect to likewise have to require the hypothesis f (∂X) ⊆ ∂Y in the global setting.
(Note that we do not want to have to make more precise assumptions such as that f (∂X) = ∂Y
or even that f preserves the index at singular points; our hope is that, as with the inverse function
theorem on open sets in sectors, the weakest necessary condition – f (∂X) ⊆ ∂Y – will turn out to
be sufficient for the desired strong conclusion on the C p property for the inverse f −1 .)
2. A simple criterion
A very basic and handy criterion for checking when a bijective C p map is a C p isomorphism is
to check if its tangent maps are isomorphisms:
Theorem 2.1. If f : X 0 → X is a bijective C p map between C p premanifolds with corners such
that f (∂X 0 ) ⊆ ∂X and the tangent map df (x0 ) : Tx0 (X 0 ) → Tf (x0 ) (X) is a linear isomorphism for
all x0 ∈ X 0 then f is a C p isomorphism.
Proof. By the inverse function theorem (which requires assuming that the map carries boundary
points into boundary points!), the hypothesis on tangent spaces implies that f is a local C p isomorphism. Thus, our problem is to show that a bijective local C p isomorphism is a C p isomorphism.
That is, we must show that the set-theoretic inverse map f −1 is a C p map. The local isomorphism
1
2
condition on the bijective f implies that for all x ∈ X there exist opens U ⊆ X around x and
U 0 ⊆ X 0 around x0 = f −1 (x) such that f restricts to a C p isomorphism between U and U 0 . In
particular, f −1 carries U into U 0 with f −1 : U → U 0 a C p map (as it must be the C p inverse to the
C p isomorphism U 0 ' U induced by f ). Since the C p property for a set-theoretic map between C p
premanifolds with corners is local on the source and target, we conclude (by varying x ∈ X) that
f −1 is indeed C p .
We now give an application of the theorem, and then explain why it is desired to impose a weaker
hypothesis. Let C ⊆ R3 be the zero locus of g(x, y, z) = x2 +y 2 −1 (a cylinder centered on the z-axis
with radius 1). For all c = (x0 , y0 , z0 ) ∈ C the functional dg(c) = 2x0 dx(c) + 2y0 dy(c) ∈ Tc (R3 )∨ is
nonzero (as x20 + y02 = 1 6= 0), so by the implicit function theorem C is a C ∞ closed hypersurface in
R3 with dimension 2, as is physically obvious. It is geometrically evident that C is a product of S 1
with R, where S 1 parameterizes the angle around the z-axis and R tracks the z-coordinates. But
we want more: C has just been given a C ∞ structure from how it sits in R3 , and so we surely expect
that the resulting bijection between C and S 1 × R is not merely a set-theoretic (or topological)
isomorphism but is even a C ∞ isomorphism (using the product C ∞ -structure on S 1 × R).
Let us now show this rigorously, as an application of the preceding theorem. The map f :
S 1 × R → R3 defined by f (θ, t) = (cos θ, sin θ, t) is C ∞ with image inside of the C ∞ embedded
submanifold C, so the induced mapping f : S 1 × R → C is C ∞ . By trigonometry, f is bijective.
A direct calculation shows that at any point ξ0 = (θ0 , t0 ) ∈ S 1 × R, the tangent map df (ξ0 ) sends
the basis vectors ∂θ |ξ0 and ∂t |ξ0 of Tξ0 (S 1 × R) to the respective vectors
− sin(θ0 )∂x |f (ξ0 ) + cos(θ0 )∂y |f (ξ0 ) , ∂z |f (ξ0 ) ∈ Tf (ξ0 ) (R3 )
that are clearly linearly independent. Hence, f is an immersion, so f is an immersion too (as f
factors through f ). But both S 1 × R and C have dimension 2 at all points, so for dimension reasons
the tangent maps of f are linear isomorphisms. Thus, by the theorem it follows that the C ∞ map
f is a C ∞ isomorphism.
Let us now explain why the isomorphism condition on tangent spaces is not always well-suited
to applications, and why we might prefer to get away with a weaker hypothesis. In the preceding
example, we had to use the geometric information that the source and target had the same dimension. That is, when proving that the tangent maps for f were injections we did not need to know
the dimension of C, but to deduce the isomorphism property we did. However, in some abstract
situations one is occasionally confronted with the situation of a C p immersion f : X → Z between
C p premanifolds such that f is bijective onto a C p subpremanifold Z0 ⊆ Z but the dimension of Z0
may not be easily computed. In such cases, the mapping properties of submanifolds ensure that f
factors through a C p map f : X → Z0 that our hypotheses assure us is bijective, and f has to be
an immersion because the immersion f factors through it. Thus, f is a bijective immersion. It is
tempting to hope that this is all we should need to know to infer that f is a C p isomorphism, at
least under some mild restrictions on X (but no real restrictions on Z0 ). The next section gives an
affirmative answer to this dream.
3. The isomorphism criterion
We now require two properties (in terms of our initial discussion above, with a bijective C p map
f : X → Y between C p premanifolds with corners such that f (∂X) ⊆ ∂Y ): X should have the
same dimension at all points (so no “jumping” occurs) and it should be second-countable (so it is
not an artificial “discrete topology” version of Y ).
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Theorem 3.1. Let f : X → Y be a C p map between non-empty C p premanifolds with corners
with 1 ≤ p ≤ ∞. Assume that f (∂X) ⊆ ∂Y , f is a bijective immersion, and X and Y are secondcountable (e.g., manifolds with corners). Also assume that X has the same dimension at all points.
The map f is then a C p isomorphism; in particular, f −1 is continuous and f (∂X) = ∂Y .
Note that in this theorem it is not assumed a priori that f preserves the index at boundary
points nor that Y has constant dimension at all points or that the dimension of each connected
component of X is equal to that of the connected component of Y into which it lands. The theorem
is one of the reasons that we require manifolds to be second-countable, and why premanifolds that
are paracompact and Hausdorff but have uncountably many connected components are considered
to be pathological. Such examples never come up in practice anyway (as one has to do something
artificial to make premanifolds with uncountably many connected components).
Proof. If {Yi } is the set of connected components of Y then these are pairwise disjoint opens covering
Y and similarly the f −1 (Yi )’s are pairwise disjoint opens covering X. It suffices to treat each of
the maps f −1 (Yi ) → Yi separately, and since each of these also satsifies the initial hypotheses on
f we may now assume that Y is connected. In particular, Y has the same dimension d ≥ 0 at all
points. The immersion property implies that each tangent space on X has dimension at most d,
so the common pointwise dimension at all points of X is at most d. Since X is second-countable,
it has only countably many connected components, say {Xn } (with n running through a subset of
the positive integers). We shall first prove that the common dimension of the Xn ’s is equal to that
of Y , and then we prove that f −1 is a C ∞ map (and so a posteriori X has to be connected).
First assume that the common dimension of the Xn ’s is some d0 < d. In this case, we claim f
cannot be surjective, contrary to hypothesis. This requires a preliminary discussion of measurezero sets in C p premanifolds with corners. Let us say that a subset S in a C p premanifold with
corners Y has measure zero if for each C p -chart (φ, U ) on Y , with φ : U → V the coordinate
map to a finite-dimensional vector space, the subset U ∩ S has measure zero with respect to the
φ-coordinates. That is, φ(U ∩ S) ⊆ V has measure zero in the usual sense. (Recall that the notion
of “measure zero” for subsets of Rn is invariant under linear change of coordinates, and so it makes
sense in abstract finite-dimensional R-vector spaces.) This is a good notion in the sense that it can
be checked with a single atlas:
Lemma 3.2. Let S be a subset of a C p premanifold with corners Y with 1 ≤ p ≤ ∞. Let {(φi , Ui )}
be a C p -atlas on Y , with φi : Ui → Σi ⊆ Vi a C p isomorphism onto an open set in a sector Σi in a
finite-dimensional R-vector space Vi . Assume φi (Ui ∩ S) has measure zero in Vi for all i. For any
C p -chart (φ, U ) on Y with φ : U → V the coordinate map, the subset φ(S ∩ U ) ⊆ V has measure
zero.
Proof. The open sets Ui ∩U cover U , but U is homeomorphic to φ(U ) ⊆ V , so U is second countable.
In particular, every open covering of U has a countable subcover, so countably many of the Ui ∩ U ’s
cover U . We may replace Y with the open subset that is the union of the corresponding countably
many Ui ’s (as this open subset of Y still contains U ), and so we can assume {Ui } is a countable
collection. Let Si = S ∩ Ui ∩ U , so φi (Si ) ⊆ φi (S ∩ Ui ) inside of Vi . In particular, φi (Si ) has measure
zero in Vi for each i.
The overlap S ∩ U is the union of its countably many subsets Si . Hence, φ(S ∩ U ) is the union of
the countably many subsets φ(Si ). Since a countable union of measure-zero sets in V has measure
zero, if each φ(Si ) has measure zero then so does φ(S ∩ U ). Thus, since φi (Si ) has measure zero
in Vi , we may focus our attention on each Si separately. That is, we fix i and rename Si as S to
get to the following special case: we have a subset S ⊆ U ∩ U 0 for C p -charts (φ, U ) and (φ0 , U 0 )
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such that φ0 (S) ⊆ V 0 has measure zero (where V 0 is the target vector space for the C p coordinate
system φ0 on U 0 ). We want to prove that φ(S) ⊆ V also has measure zero. We may assume U ∩ U 0
is non-empty, as otherwise S is empty and the problem is trivial. Hence, dim V = dim V 0 , say this
common dimension is n.
By the compatibility condition on all C p -charts for a common C p -structure, the transition map
φ ◦ φ0 −1 is a C p -isomorphism from the open subset φ0 (U ∩ U 0 ) ⊆ Σ0 ⊆ V 0 onto the open subset
φ(U ∩ U 0 ) ⊆ Σ ⊆ V (with Σ and Σ0 sectors in V and V 0 ). This map carries φ0 (S) over to φ(S), so
we come down to the following rather concrete question. Consider n-dimensional vector spaces V
and V 0 over R and a C p isomorphism f : A0 ' A between open subsets A ⊆ Σ and A0 ⊆ Σ0 where
Σ ⊆ V and Σ0 ⊆ V 0 are sectors. For a subset S ⊆ A0 , we claim that S has measure zero in V 0 if
and only if f (S) has measure zero in V . It suffices to prove that f (S) has measure zero whenever
S has measure zero, as such a general assertion may then be applied to the C p isomorphism f −1
in the other direction to deduce the converse. We may pick linear coordinates on V and V 0 , so our
problem becomes: if f is a C p isomorphism between open sets in sectors in Rn , with 1 ≤ p ≤ ∞,
then f carries measure-zero sets to measure-zero sets. This follows from the lemma below.
Lemma 3.3. If Σ ⊆ Rn is a sector and f : U → Rn is a C 1 mapping on an open set U ⊆ Σ then
for any measure zero set A ⊆ U the image f (A) ⊆ Rn has measure zero.
Proof. We may make a linear change of coordinates on the source so that Σ is a “standard” sector
(namely, Rn or [0, ∞)r × Rn−r for some 1 ≤ r ≤ n). In this setting, the old proof of the result (in
the case Σ = Rn ) carries over verbatim. One can also use the local version of Whitney’s extension
theorem in conjunction with a countability argument to reduce the problem on sectors to the case of
open sets in Rn , and we leave it to the interested reader to work out such an alternative argument
(this it is overkill to do this, since the old argument really works on sectors).
We now make a definition:
Definition 3.4. For 1 ≤ p ≤ ∞, a subset S in a C p premanifold with corners Y has measure zero
if for some C p -atlas {(φi , Ui )} covering Y , each φi (S ∩ Ui ) has measure zero in the target vector
space for φi .
The preceding lemma ensures that the condition in this definition is satisfied for all C p -atlases
on Y if it is satisfied for one. Obviously a measure-zero set in a C p premanifold with corners cannot
fill up the whole space, since even in the domain of any C p chart it cannot fill up that domain (as a
non-empty open set in a sector in a finite-dimensional R-vector space does not have measure zero).
Also, a countable union of measure zero sets in Y has measure zero. Indeed, by definition we may
check this in local C p -charts, and thus the problem is shifted into open sets in sectors in Rn ; it is
clear that countable unions of measure zero sets in Rn have measure zero.
As a consequence of these consideration with sets of measure zero, we may now return to our
original problem of proving that a bijective C p -immersion f : X → Y between second countable C p
premanifolds with corners is a C p isomorphism when X has the same dimension at all of its points
and f (∂X) ⊆ ∂Y . We have already reduced to the case when Y is connected, say with dimension
d, and so each of the countably many connected components Xn of X has a common dimension
d0 ≤ d. The key is:
Lemma 3.5. With notation as above, necessarily d0 = d.
Proof. Suppose otherwise, so each Xn has dimension < d. If we can show that f (Xn ) ⊆ Y has
measure zero in Y for all n, then the countable union f (X) of the f (Xn )’s would have measure zero
in Y . But this would force f (X) to be a proper subset of Y , contradicting the assumption that f
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is bijective. To show that each f (Xn ) has measure zero in Y , we note that since the topological
space Xn is a second-countable space, any open cover has a countable subcover. Likewise, Y has
the same property. Let {(φm , Um )} be a covering of Y by countably many C p -charts, so each open
0 )}
f −1 (Um ) in Xn is covered by countably many C p -charts {(ψr,m , Ur,m
r≥1 . Since f (Xn ) is the union
0
of the countably many sets f (Ur,m ), it suffices to show that each of these has measure zero. Clearly
0 ) is contained in U , so upon renaming U as Y and U 0
f (Ur,m
m
m
r,m as X we may assume that X is
0
0
d
an open set in a sector Σ in R and Y is an open set in a sector Σ in Rd , and that we are given
a C p map from X to Y . In this case, since d0 < d the image f (X) has measure zero in Rd . The
reason is as follows. The map f may be factored as the composite of the inclusion of X into the
0
0
open set U = X × Rd−d in the sector Σ0 × Rd−d in Rd via x 7→ (x, 0) (where its image in Rd is a
measure zero set, as d0 < d) followed by the C p map U → Rd defined by (x, y) 7→ f (x) + y. Thus,
we get the desired result by Lemma 3.3.
We have now proved that the constant dimension of X must equal that of Y . Hence, the
injective tangent maps df (x) : Tx (X) → Tf (x) (Y ) must be linear isomorphisms for all x ∈ X, so
since f (∂X) ⊆ ∂Y the inverse function theorem “with corners” may be used to conclude that the
map f is a local C p isomorphism. However, f is bijective, so we conclude that its inverse map f −1
is a local C p isomorphism as well. Thus, f −1 is a C p mapping and hence f is a C p isomorphism.
This concludes the proof of the theorem.
4. Applications
We can use Theorem 3.1 to prove uniqueness of differentiable structures on various subsets of
manifolds.
Let X be a C p premanifold with corners (such as a C ∞ manifold with boundary), with 1 ≤ p ≤ ∞,
and for r ≥ 0 let Xr be the locally closed subset of points on X with index r. In §3 in the handout
on premanifolds with corners it was shown how to give the subset Xr a structure of C p premanifold
so that it has a universal mapping property: for any C p premanifold with corners X 0 and a C p
map f : X 0 → X whose image is contained in Xr , the resulting map of sets f : X 0 → Xr (which
is continuous, as Xr has the subspace topology from X) is a C p mapping. We can now exploit
this property to prove much more. First, the C p -structure that we have constructed on Xr has an
extra property: by construction, it makes the inclusion mapping ir : Xr → X a C p mapping that
is moreover an immersion: dir (x) : Tx (Xr ) → Tx (X) is injective on tangent spaces for all x ∈ Xr .
(This basically amounts to the fact that we may take local C p coordinates {y1 , . . . , ym } on Xr
around x to be the restriction to U ∩ Xr of the first m members in a system of local C p coordinates
{ξ1 , . . . , ξM } on an open U around x in X, and dir (x) takes ∂yj |x to ∂ξi |x .) This provides the input
to prove a strong uniqueness theorem that may help to convince the reader that the C p structure
we have put on Xr is not ad hoc:
Theorem 4.1. The C p -premanifold structure we have on Xr is the unique one with respect to
which the inclusion Xr → X is an immersion.
We emphasize a special case: if X is a C p premanifold with boundary, then the C p structure we
have put on the closed subset ∂X = X1 is the unique one with respect to which the inclusion map
∂X → X is an immersion. This midly generalizes Exercise 12 in Chapter 2 of the course text.
Proof. Let Xr0 denote Xr equipped with another such structure, so by hypotheses we have an
injective immersion i0r : Xr0 → X. By the mapping property for ir : Xr → X using the “usual” C p
structure, we get a unique factorization of i0r as i0r = ir ◦ h for a C p mapping h : Xr0 → Xr . The
map h must be the identity on underlying topological spaces, and since i0r and ir are immersions
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it follows from the Chain Rule that h is an immersion. Hence, the identity map Xr0 → Xr is an
immersion. The equality of C p structures is exactly the claim that the identity map on underlying
spaces is a C p isomorphism. Hence, our problem is now a special case of the following general
claim: if h : M 0 → M is a C p immersion between C p premanifolds and h is a homeomorphism,
prove that it is a C p isomorphism.
This problem is local over M (i.e., if {Ui } is an open covering of M then the restricted maps
h−1 (Ui ) → Ui have the same properties as h, and these are C p isomorphisms if and only if h is),
so we can replace M with a local chart to ensure that M is second countable and Hausdorff (i.e., a
manifold). Since h is a homeomorphism, this forces M 0 to also be second countable and Hausdorff.
We may also work separately over the (open) connected components of M , so we can assume that
M is connected and hence (since h is a homeomorphism) M 0 is also connected. Thus, the pointwise
dimensions on M and M 0 are constant. By Theorem 3.1, we may now conclude that h is a C p
isomorphism.
Now let X be a C p premanifold (no corners!), 1 ≤ p ≤ ∞, and suppose that X is equipped with
a properly discontinuous action of a discrete group G. We therefore get a C p premanifold quotient
map π : X → X/G that is a G-invariant local C p isomorphism and satisfies a universal mapping
property for G-invariant C p maps from X to other C p premanifolds. We want to give a simple
criterion to show that a concretely-given G-invariant surjective C p map X → X 0 “is” the quotient
map X → X/G in disguise.
In view of what has gone above, we shall have to assume that X is second-countable. Note
that this forces X/G to be second-countable (as follows immediately from the construction of
the topology on X/G, with all sufficiently small opens obtained as images of opens in X). This
motivates the hypotheses is:
Theorem 4.2. For X and G as above, in particular with X second countable, suppose that the
map f : X → X 0 is a surjective C p map onto a second-countable C p premanifold X 0 such that
f (x.g) = f (x) for all x ∈ X and g ∈ G. Assume also that X has constant pointwise dimension
and that each fiber f −1 (x0 ) is a G-orbit in X. If f is an immersion then the induced C p map
f : X/G → X 0 is a C p isomorphism. In particular, f : X → X 0 “is” the C p premanifold quotient
of X by the action of G.
This theorem gives a concrete condition for identifying a C p surjection f : X → X 0 as the
premanifold quotient by the G-action: f should be an immersion and the the fibers should be
G-orbits. Specific examples will be given after the proof.
Proof. Since X → X/G induces isomorphisms on tangent spaces, the induced C p map f : X/G →
X 0 is an immersion and X/G has constant pointwise dimension (the same as that of X). Also,
X/G is second countable (since X is), and f is surjective because f is. Since it is assumed that the
fibers of f are G-orbits, it follows that f is also injective and hence bijective. By Theorem 3.1, f
is therefore a C p isomorphism.
We now apply this theorem to efficiently handle some earlier problems that we had concerning
the comparison of multiple C p -structures on the same topological space.
Example 4.3. The map R → S 1 ⊆ R2 given by t 7→ (cos(2πt), sin(2πt)) is a C ∞ map from R to
R2 , and since S 1 is a closed submanifold of the plane it follows that this map factors through a C ∞
map R → S 1 that is clearly invariant with respect to the Z-action on R by additive translation. By
trigonmetry, this map is surjective and its fibers are Z-orbits, so the induced C ∞ map R/Z → S 1
is bijective and hence (by Theorem 3.1) is a C ∞ isomorphism. This says that the C ∞ -structure
on S 1 through planar coordinates is the same as through polar coordinates, but it avoids the need
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to do explicit calculations: we let the properties of the trigonometric functions do all of the work.
(These properties are hiding in the fact that f is bijective and f is a C ∞ map from R to the plane.)
Example 4.4. Let us revisit the problem of showing that the C ∞ -structures on Pn (R) through its
standard atlas of n + 1 Euclidean spaces and through its presentation as the antipodal quotient
of the standard n-sphere S n ⊆ Rn+1 are in fact the same C ∞ -structure. This was proved in an
earlier handout by explicit computations with charts. We now do it with essentially no explicit
computations, letting the geometry of tangent spaces and an intelligent choice of coordinates at
the end do all of the work.
By Theorem 4.2, it suffices to show that when Pn (R) is given the C ∞ structure arising from
the atlas of Euclidean charts, then the natural topological quotient map S n → Pn (R) (which is
surjective with antipodal pairs as fibers) is an immersion. Let V = Rn+1 with its standard inner
product, so Pn (R) = P(V ) and S n is identified with the unit sphere S in V ∨ . The quotient map
S n → Pn (R) is then identified with the composite of the C ∞ embedding ι : S → V ∨ − {0} and
the submersive C ∞ map F : V ∨ − {0} → P(V ) that was studied in an earlier homework. (Settheoretically, F sends a functional ` ∈ V ∨ − {0} to the hyperplane ker ` as a point of P(V ).) This
shows that the map S → P(V ) is C ∞ with respect to the C ∞ -structure we have put on P(V ).
To check the immersion criterion in Theorem 4.2, by the Chain Rule we just have to make sure
that for each s ∈ S the hyperplane Ts (S) inside of Ts (V ∨ − {0}) has trivial intersection with the
kernel line of the surjective map
dF (s) : Ts (V ∨ − {0}) → TF (s) (P(V )).
(Indeed, this implies that d(F ◦ ι)(s) is injective for all s ∈ S, as desired.) Let L ⊆ V ∨ be the line
spanned by the nonzero point s, so L−{0} in V ∨ is a punctured line passing through s whose image
in P(V ) under F is the single point F (s) (as all nonzero multiples of s are nonzero scalar multiples
of a common functional on V and hence all have the same kernel hyperplane in V ). Since the map
L − {0} → P(V ) factors through a point, its induced map on tangent spaces must be zero because
(by the Chain Rule) it factors through the vanishing tangent space of a 1-point manifold. Thus,
Ts (L − {0}) is a line in the kernel of the surjection dF (s), yet for dimension reasons this surjective
map has 1-dimensional kernel. We conclude that ker dF (s) = Ts (L − {0}), and so our problem is
to check that the subspaces Ts (S) and Ts (L − {0}) in Ts (V ∨ − {0}) have trivial intersection. This
is a problem that is internal to V ∨ : the projective space P(V ) has now dropped out of the picture.
Renaming V ∨ as W , our problem is to show that if W is a finite-dimensional inner product space,
s is a point on the unit sphere S in W , and L ⊆ W is the line spanned by s 6= 0, the subspaces
Ts (S) and Ts (L) have vanishing intersection inside of Ts (W ). We may now choose an orthonormal
P
basis e = {ei } of W with e1 = s, so W = Rm with the standard inner product, S = { x2i = 1},
s = (1, 0, . . . , 0), and L is the x1 -axis (given by x2 = · · · = xn = 0). In Ts (W ) withPits standard
basis {∂xi |s }, the line Ts (L) is the span of ∂x1 |s . Since S is the zero locus of f =
x2i − 1, the
hyperplane Ts (S) is the kernel of the functional
X
X
df (s) :
ai ∂xi |s 7→
2xi (s)ai = 2a1
on Ts (W ). Hence, Ts (S) is spanned by the ∂xi |s ’s for i > 1, so it indeed has trivial intersection
with the line Ts (L) that is spanned by ∂x1 |s .